Question

Keely Temp Services Agency finds it difficult to retain its employees because most of them are looking for full-time positions and will leave the Keely Agency when a good opportunity comes along. The Agency’s records show that the average stay of its employees was 6 months for the last several years. In an effort to reduce the cost associated with this high level of employee turnover, the management of the company decided to introduce new incentives to encourage employees to stay longer. Their goal is to try to increase the average stay of the Agency’s employees to at least 10 months. For 60 most recent Keely employees who terminated after the incentive measures were introduced, the average time of employment was 7.2 months. Assume that this is a random sample of Keely employees and the population standard deviation for the duration of employment of Keely employees is 10.5 months. At 1% level of significance, do you think the Agency is achieving its goal? Please show all the necessary steps and explain your conclusion about the hypothesis.  

Answer 1

Solution:

Here, we have to use one sample z test for the population mean.

The null and alternative hypotheses are given as below:

Null hypothesis: H0: the average stays of the Agency’s employees is at least 10 months.

Alternative hypothesis: Ha: the average stays of the Agency’s employees is less than 10 months.

H0: µ ≥ 10 versus Ha: µ < 10

This is a lower tailed test.

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 10

Xbar = 7.2

σ = 10.5

n = 60

α = 0.01

Critical value = -2.3263

(by using z-table or excel)

Z = (7.2 – 10)/[10.5/sqrt(60)]

Z = -2.0656

P-value = 0.0194

(by using Z-table)

P-value > α = 0.01

So, we do not reject the null hypothesis

There is sufficient evidence to conclude that the average stays of the Agency’s employees is at least 10 months.